【全程復(fù)習(xí)方略】2020年人教A版數(shù)學(xué)理(廣東用)課時(shí)作業(yè)：第二章-第一節(jié)函數(shù)及其表示

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)提升作業(yè)(四)一、選擇題1.(2022·江西高考)若函數(shù)f(x)=QUOTE則f(f(10))=()(A)lg101 (B)2 (C)1 (D)02.(2021·中山模擬)下列各組函數(shù)中表示同一個(gè)函數(shù)的是()(A)f(x)=QUOTE,g(x)=QUOTE(B)f(x)=QUOTE·QUOTE,g(x)=QUOTE(C)f(x)=QUOTE,g(x)=x0(D)f(x)=QUOTE,g(x)=x-13.(2021·廣州模擬)函數(shù)y=QUOTE的定義域?yàn)?)(A)(QUOTE,1) (B)(QUOTE,+∞)(C)(1,+∞) (D)(QUOTE,1)∪(1,+∞)4.設(shè)f(x)=QUOTE則f(5)的值為()(A)10 (B)11 (C)12 (D)135.函數(shù)f(x)=QUOTE+lgQUOTE的定義域是()(A)(2,4) (B)(3,4)(C)(2,3)∪(3,4] (D)[2,3)∪(3,4)6.假如f(QUOTE)=QUOTE,則當(dāng)x≠0且x≠1時(shí),f(x)=()(A)QUOTE (B)QUOTE (C)QUOTE (D)QUOTE-17.(2021·惠州模擬)已知函數(shù)f(x)=QUOTE若f(a)=2,則a=()(A)4 (B)2 (C)1 (D)-18.函數(shù)f(x)=QUOTE(x≠-QUOTE)滿足f(f(x))=x,則常數(shù)c等于()(A)3 (B)-3(C)3或-3 (D)5或-39.已知函數(shù)y=f(x+1)的定義域是[-2,3],則y=f(2x-1)的定義域是()(A)[0,QUOTE] (B)[-1,4](C)[-5,5] (D)[-3,7]10.(力氣挑戰(zhàn)題)已知函數(shù)y=f(x)的圖象關(guān)于直線x=-1對(duì)稱,且當(dāng)x∈(0,+∞)時(shí),有f(x)=QUOTE,則當(dāng)x∈(-∞,-2)時(shí),f(x)的解析式為()(A)f(x)=-QUOTE (B)f(x)=-QUOTE(C)f(x)=QUOTE (D)f(x)=-QUOTE二、填空題11.已知兩個(gè)函數(shù)f(x)和g(x)的定義域和值域都是集合{1,2,3},其函數(shù)對(duì)應(yīng)關(guān)系如下表:x123f(x)231x123g(x)321則方程g(f(x))=x的解集為.12.(2021·石家莊模擬)已知函數(shù)f(x)=QUOTE若f(f(0))=4a,則實(shí)數(shù)a=.13.二次函數(shù)的圖象經(jīng)過(guò)三點(diǎn)A(QUOTE,QUOTE),B(-1,3),C(2,3),則這個(gè)二次函數(shù)的解析式為.14.函數(shù)y=lg(ax2-2ax+2)的定義域?yàn)镽,則a的取值范圍是.三、解答題15.(力氣挑戰(zhàn)題)假如對(duì)任意實(shí)數(shù)x,y,都有f(x+y)=f(x)·f(y),且f(1)=2,(1)求f(2),f(3),f(4)的值.(2)求QUOTE+QUOTE+QUOTE+…+QUOTE+QUOTE+QUOTE的值.答案解析1.【解析】選B.∵f(10)=lg10=1,∴f(f(10))=f(1)=12+1=2.2.【解析】選C.對(duì)于A,f(x)的值域大于等于0,而g(x)的值域?yàn)镽,所以A不對(duì);對(duì)于B,f(x)的定義域?yàn)閧x|x≥1};而函數(shù)g(x)的定義域?yàn)閧x|x≥1或x≤-1},所以B不對(duì);對(duì)于C,由于f(x)=QUOTE=1(x≠0),g(x)=x0=1(x≠0),所以兩個(gè)函數(shù)是同一個(gè)函數(shù),所以C對(duì);對(duì)于D,f(x)的定義域?yàn)閧x|x≠-1};而函數(shù)g(x)的定義域?yàn)镽,所以D不對(duì).3.【解析】選A.要使函數(shù)有意義,則QUOTE即QUOTE∴QUOTE<x<1,∴函數(shù)的定義域?yàn)?QUOTE,1).4.【解析】選B.f(5)=f(f(11))=f(9)=f(f(15))=f(13)=11.【方法技巧】求函數(shù)值的四種類型及解法(1)f(g(x))型:遵循先內(nèi)后外的原則.(2)分段函數(shù)型:依據(jù)自變量值所在區(qū)間對(duì)應(yīng)求值,不確定時(shí)要分類爭(zhēng)辯.(3)已知函數(shù)性質(zhì)型:對(duì)具有奇偶性、周期性、對(duì)稱性的函數(shù)求值,要用好其函數(shù)性質(zhì),將待求值調(diào)整到已知區(qū)間上求解.(4)抽象函數(shù)型:對(duì)于抽象函數(shù)求函數(shù)值,要用好抽象的函數(shù)關(guān)系,適當(dāng)賦值,從而求得待求函數(shù)值.5.【解析】選D.要使函數(shù)有意義,必需QUOTE所以函數(shù)的定義域?yàn)閇2,3)∪(3,4).6.【解析】選B.令QUOTE=t,t≠0且t≠1,則x=QUOTE,∵f(QUOTE)=QUOTE,∴f(t)=QUOTE,化簡(jiǎn)得:f(t)=QUOTE,即f(x)=QUOTE(x≠0且x≠1).7.【解析】選A.當(dāng)a>0時(shí),由log2a=2得a=4;當(dāng)a≤0時(shí),由a+1=2得a=1,不合題意,舍去,故a=4.8.【解析】選B.f(f(x))=QUOTE=x,∴f(x)=QUOTE=QUOTE,得c=-3.9.【解析】選A.由-2≤x≤3,得-1≤x+1≤4,由-1≤2x-1≤4,得0≤x≤QUOTE,故函數(shù)y=f(2x-1)的定義域?yàn)閇0,QUOTE].10.【思路點(diǎn)撥】函數(shù)y=f(x)的圖象關(guān)于直線x=-1對(duì)稱,則有f(x)=f(-x-2).【解析】選D.設(shè)x<-2,則-x-2>0,由函數(shù)y=f(x)的圖象關(guān)于x=-1對(duì)稱,得f(x)=f(-x-2)=QUOTE,所以f(x)=-QUOTE.11.【解析】當(dāng)x=1時(shí),f(x)=2,g(f(x))=2,不合題意;當(dāng)x=2時(shí),f(x)=3,g(f(x))=1,不合題意;當(dāng)x=3時(shí),f(x)=1,g(f(x))=3,符合要求,故方程g(f(x))=x的解集為{3}.答案:{3}12.【解析】∵f(0)=20+1=2,∴f(f(0))=f(2)=4+2a=4a,∴a=2.答案:213.【解析】方法一:設(shè)y-3=a(x+1)(x-2),把A(QUOTE,QUOTE)代入得a=1,∴二次函數(shù)的解析式為y=x2-x+1.方法二:設(shè)二次函數(shù)解析式為y=ax2+bx+c,則有QUOTE解得QUOTE∴二次函數(shù)的解析式為y=x2-x+1.答案:y=x2-x+114.【解析】當(dāng)a=0時(shí),函數(shù)為y=lg2,定義域?yàn)镽滿足題意.當(dāng)a≠0時(shí),要使函數(shù)y=lg(ax2-2ax+2)的定義域?yàn)镽,必需QUOTE即QUOTE解得0<a<2.故a的取值范圍為[0,2).答案:[0,2)15.【思路點(diǎn)撥】(1)依據(jù)等式中變量的任意性,可接受賦值法求函數(shù)值.(2)依據(jù)(1)的函數(shù)值相鄰兩項(xiàng)的規(guī)律求出比值,然后求解.【解析】(1)∵對(duì)任意實(shí)數(shù)x,y,都有f(x+y)=f(x)·f(y),且f(1)=2,∴f(2)=f(1+1)=f(1)·f(1)=22=4,f(3)=f(2+1)=f(2)·f(1)=23=8,f(4)=f(3+1)=f(3)·f(1)=24=16.(2)由(1)知QUOTE=2,QUOTE=2,QUOTE=2,…,QUOTE=2.故原式=2×1007=2022.【變式備選】已知a